The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.
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The relationship of ${\frak m+m=m}$ to AC
Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.)
Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$?
If not, what is the ...
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1answer
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(Non) equivalence of regular cardinal definitions
The usual definition of a regular cardinal is "$\kappa$ is regular if $cf(\kappa) = \kappa$", which, assuming the axiom of choice, is equivalent to this definition: "$\kappa$ is regular iff it cannot ...
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Zorn's lemma and maximal linearly ordered subsets
Let $T$ be a partially ordered set. We say $T$ is a tree if $\forall t\in T$ $\{r\in T\mid r < t\}$ is linearly ordered (such orders can be considered on connected graphs without cycles, i.e. on ...
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Axiom of Choice-esque argument to show that a proof of a statement exists without actually giving a proof
What if the set of all well-formed statements in ZFC formed a kind of pseudo-category where a morphism f between objects A, B represented a formal proof that A implied B? What if that category could ...
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1answer
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Jech: Set Theory exercise 3.13, how do I avoid Choice?
In an effort to finally learn set theory rigourously, I've decided to start plowing through Jech's Set theory, making sure to do each of the exercises.
Here is Jech's problem 3.13 (pg. 34 in the ...
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1answer
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Can we write every uncountable set $U$ as $V∪W$, where $V$ and $W$ are disjoint uncountable subsets of $U$? [duplicate]
Is it true that for every uncountable set $U$, we can write $U=V∪W$, where $V$ and $W$ are disjoint uncountable subsets of $U$ ?
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Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?
The answer to Cardinality of a locally compact space without isolated point shows that AC is required to show that if $X$ is a compact Hausdorff space with no isolated points then $|X| \ge ...
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6answers
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Can every infinite set be divided into pairwise disjoint subsets of size $n\in\mathbb{N}$?
Let $S$ be an infinite set and $n$ be a natural number. Does there exist partition of $S$ in which each subset has size $n$?
This is pretty easy to do for countable sets. Is it true for ...
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1answer
48 views
Dissecting a proof of the $\Delta$-system lemma (part II)
This is part II of this question I asked yesterday. In the link you can find a proof of the $\Delta$-system lemma. In case 1 it uses the axiom of choice (correct me if I'm wrong). Now one can also ...
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2answers
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Isomorphic Free Groups and the Axiom of Choice
When I read about free group, the proof which concerns about two free groups $F(X)$ and $F(Y)$ are isomorphic only if $\operatorname{card}(X) = \operatorname{card}(Y)$ has a sentence going as follows:
...
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Understanding a proof of Diaconescu's theorem
I am trying to walk through the proof of Diaconescu's theorem that the axiom of choice implies the law of excluded middle at http://plato.stanford.edu/entries/intuitionism/#ChoAxi. To paraphrase:
...
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1answer
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Dependent choice and Zorn's Lemma
How much of Zorn's lemma can be saved if we assume only ZF+DC without full choice?
More precisely: assume we have a partially ordered (inductive) set which is of size continuum. Then can we apply ...
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2answers
127 views
The Axiom of Choice and definability
I've seen a lot of relations between the notion of the existence of a definable set with a given property and the necessity of AC is proving that there is a set with the property. For example:
Under ...
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1answer
70 views
Tychonoff Theorem and the axiom of choice
How to show that
The Tychonoff Theorem and the axiom of choice are equivalent?
Here I want to collect ways to prove it.
Thanks for your help.
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2answers
563 views
Predicting Real Numbers
Here is an astounding riddle that at first seems impossible to solve. I'm certain the axiom of choice is required in any solution, and I have an outline of one possible solution, but would like to ...
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1answer
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Cauchy functional equation with non choice
Assume ZF+ not AC. Then how many solutions are there for Cauchy functional equation?
Thank you
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2answers
40 views
Mapping on cardinal without Axiom of Choice
Define $|A|\le|B|$ iff there exists injective mapping $A \to B$.
If Axiom of Choice is assumed then this is equivalent as $|A|\le|B|$ iff there exists surjective mapping $B \to A$. But:
If Axiom of ...
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1answer
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Does a nonlinear additive function on R imply a Hamel basis of R?
A function is additive if $f(x+y) = f(x) + f(y)$. Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$. But assuming the axiom of ...
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2answers
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Intuition behind the Axiom of Choice
Why is it different to make one choice or many choices than to make infinite choices from a theoretical point of view in which indeed you are not going to do any?
How could that be different from ...
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3answers
161 views
Why the need of Axiom of Countable Choice?
Two theorems:
$(1)$ Countable Union of Countable Sets is Countable
$(2)$ Cartesian Product of Countable Sets is Countable
Linked are the formal proofs on Proofwiki.
I do not understand why they ...
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votes
3answers
81 views
$\mathbb{R^+}$ is the disjoint union of two nonempty sets, each closed under addition.
I saw Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition. and have a question related to the answer (I'm not sure if this is the right place to post ...
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2answers
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Are there versions of the axiom of choice that restrict the size of the factors?
One formulation of the axiom of choice is that an arbitrary product of nonempty sets must be nonempty. The axiom of countable choice AC$_\omega$ is known to be strictly weaker than AC, but still ...
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1answer
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Stone's Representation Theorem and The Compactness Theorem
If you're working on $ZF$ and you assume the compactness theorem for propositional logic, then you have the prime ideal theorem, and thus you can show that the dual of the category of Boolean algebras ...
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2answers
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Does the existence of this quotient set depend on the Axiom of Choice?
We know the familiar equivalent relation on $\mathbb{R}$, which is
$$
x\sim y\Leftrightarrow x-y\in\mathbb{Q}
$$
After quoting this relation, we have the quotient set
$$
\mathbb{R}/_\sim = \{x + ...
3
votes
1answer
60 views
well-ordering principle
I'm confused by the well-ordering principle . The proof is clear but I don't have any idea why is it true . It says that every non-empty set can be well-ordered but $C^{1}$ is a non-empty set but ...
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3answers
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Proof of $CFE \implies BPI$
(CFE): Every filter of closed sets can be extended to a maximal one.
(BPI): Every Boolean algebra contains a prime ideal.
I am reading Herrlich's and Stepran's paper "Maximal filters, continuity and ...
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2answers
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Question about proof in Jech's The axiom of choice ($AC_\omega$)
I'm looking at the following in Jech's The Axiom of Choice on page 20:
2.4.1. Example: The Countable Axiom of Choice implies that every infinite set has a countable subset.
Proof. Let $S$ be ...
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1answer
88 views
How to prove that a regular cardinal cannot be expressed as a union of sets with less cardinality?
My question is about the following:
Using the Axiom of Choice show that:
If $\kappa\ge\omega$ is a regular cardinal, $\gamma\le\kappa$, and $\langle A_\alpha\mid\alpha\lt\gamma\rangle$ is a ...
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votes
4answers
114 views
Do we always use the Axiom of Choice when picking from uncountable number of sets?
I know there are uncountable number of equivalence classes defined by the relation defined here: Equivalence classes of "$x \sim y \Longleftrightarrow x -y $ is rational". (the relation is ...
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1answer
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Proof of a basic $AC_\omega$ equivalence
On Wikipedia it is mentioned that "... in order to prove that every accumulation point $x$ of a set $S\subseteq \mathbf R$ is the limit of some sequence of elements of $S\setminus \{x\}$, one uses (a ...
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Axiom of Choice and Determinacy
In my set theory course we have been talking about the axiom of determinacy. One of the first things we showed was that $AD$ and $AC$ are incompatible. We later showed that $ZF+AD$ implies the ...
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3answers
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Equivalence with the axiom of choice
This is a problem from Tao's Analysis I.
We are asked to show that the axiom of choice is equivalent to the statement that for any sets $A$ and $B$ for which a surjection $g:B\to A$ exists, an ...
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1answer
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Failure of Choice only for sets above a certain rank
Let $\alpha$ be an ordinal. How can we show that the following theory is consistent?
$\mathrm{ZF}$ + "there exists a set with rank greater than $\alpha$ that is not well ordered" + "every set of rank ...
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3answers
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Why does the infinite prisoners and hats puzzle require the axiom of choice?
Infinite prisoners puzzle.
The link to Wikipedia describes the puzzle, and the solution. The axiom of choice is used to pick a sequence from each equivalence class, which the prisoners memorize ...
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1answer
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Axiom of choice , Hartogs ordinals, well-ordering principle
I'm trying to prove the following:
If it holds that if for any two sets $A$ and $B$, $A$ can be injected into $B$ or $B$ can be injected into $A$, then every set can be well-ordered (axiom of choice ...
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1answer
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Can one construct an algebraic closure of fields like $\mathbb{F}_p(T)$ without Zorn's lemma?
I have heard that an algebraic closure of $\mathbb{Q}$ can be constructed without Zorn's lemma and so can an algebraic closure of a finite field $\mathbb{F}_p$. What about $\mathbb{F}_p(T)$? Do there ...
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1answer
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A union represented as a disjoint union: weaker than choice?
While looking at this problem, I was thinking about the more general statement:
For all sets $I,X$ and $U:I\to \mathcal P(X)$, the power set of $X$, there exists a $V:I\to \mathcal P(X)$ with the ...
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3answers
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A question about a proof in one of Sierpiński's papers
The following is a question about Sierpiński's paper "Une démonstration du théorème sur la structure des ensembles de points", (link):
We call a set dense-in-itself if it does not contain any ...
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1answer
159 views
Do you need the Axiom of Choice to assert that every real vector space has a norm?
Math people:
This question is 95% answered (the first answer) at Does every $\mathbb{R},\mathbb{C}$ vector space have a norm? and Vector Spaces and AC . The questions, answers, and links found there ...
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1answer
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Proving that a dense set in a poset by Zorn's lemma have a maximal antichain as a subset
So I am little confused here; Zorn's lemma says that if every chain has a supremum in a poset one defines, then the poset must have a maximal element. My question is, how does this lead to the proof ...
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1answer
48 views
Does $n^A\cong n^B \Rightarrow A\approx B$ require choice?
Let $V$ be a vector space over a finite field $F$.
Assume $V$ has a basis $S$.
Then, define $\Phi:V^*\rightarrow F^S:f\mapsto f\upharpoonright S$. It can be shown that $\Phi$ is an isomorphism. Thus, ...
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1answer
103 views
Injection of union into disjoint union
Given a family of sets $(A_i : i \in I)$, we define the disjoint union:
$$\sum_{i \in I} A_i = \bigcup_{i \in I} (\{i\} \times A_i).$$
There is a surjection $\sum_{i \in I}A_i \to \bigcup_{i \in I} ...
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1answer
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Clarification of a proof in Herrlich
In Herrlich on page 5 he gives a proof of $\textbf{AC} \implies \textbf{WOT}$:
He does not give a definition of cardinality $|X|$ before this proof and I searched the index for a definition but ...
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4answers
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Cardinals of set operations without AC
Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in ZF (no axiom of choice).
Prove: $|A\cup B|=\mathfrak{c}$
If $B \subset A\implies|A \backslash B|=\mathfrak{c}$?
I have found several places ...
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2answers
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Wellordering vs. Zorn's lemma
Many mathematicians outside mathematical logic dislike wellorderings, ordinals and corresponding transfinite arguments. They use zorn's lemma instead and claim one does not need ordinals at all. ...
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Is Dover publishing Moore's book on the Axiom of Choice? [closed]
Dover is publishing a paperback edition of Gregory H. Moore's Zermelo's Axiom of Choice: Its Origins, Development, and Influence. It's supposed to come out March 20th and is available for pre-order at ...
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A confusion about Axiom of Choice and existence of maximal ideals.
The proof of the theorem of statement that every ring has a maximal ideal uses zorn's lemma or the axiom of choice.Now, the defintion of ring as well as the definition of maximal ideal don't depend on ...
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2answers
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Application of Zorn's Lemma
I have the following problem:
Given $\{A_{\lambda}\}_{\lambda \in \Lambda}$ for some index set $\Lambda$. Prove that there exists $\{B_{\lambda}\}_{\lambda \in \Lambda}$ such that:
1) $(\forall ...
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The “it's not possible” statement in math and the Axiom of Choice
This question actually consists 3 related pieces of text, which I've gathered under this title about which I would like your opinion (they rather contain the implicit question "is this the right way ...
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1answer
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Is every discrete topological space orderable?
I apologize for asking a question in topological terms when it's not really about topology, but here goes:
If $S$ is a set, is it always possible to totally order $S$ so that every non-minimal ...
